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4 years ago

What star along the main sequence will likely end in a supernova

Physics

2 answers:

snow_tiger [21]4 years ago

7 0

Blue-White stars will likely end in a supernova.

MrMuchimi4 years ago

4 0

Don't know $\sqrt{x} \frac{x}{y} \frac{x}{y} \frac{x}{y} \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$

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The frequency is slowly increased. Once it passes the critical value fb, the student hears the ball bounce. There is now enough

vitfil [10]

Answer: g = acceleration = A*w^2 = A*(2*pi*fb)^2.

Explanation:

The ball bounces when the acceleration of the ball exceeds that of gravity. If A and fb are measured at that point, g = acceleration = A*w^2 = A*(2*pi*fb)^2.

3 0

3 years ago

A calculator has a resistance of 22 Ω. What is the power rating for this calculator when connected to a 1.5 V battery?

GREYUIT [131]

Answer:

Explanation:

3 0

3 years ago

Rain and wind place which type of load on structures?

katovenus [111]

It’s a dynamic load.
Hopes this helps :)

4 0

3 years ago

Read 2 more answers

Look at the diagram provided. What would be the mechanical energy at Point #4?

goblinko [34]

Answer:

18000 J

Explanation:

From the question given above, the following data were obtained:

At point 4:

Mass of cart = 600 Kg

Velocity of cart (v) = 7.745 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 600 × 7.745²

KE = 300 × 7.745²

KE ≈ 18000 J

Therefore, the mechanical energy of the cart at point 4 is 18000 J

6 0

3 years ago

. You have completed your design of a new surgical headlamp. You have decided to power the headlamp using four AA rechargeable b

SSSSS [86.1K]

Answer:

LED run before the batteries are depleted is 3.87 hours

Explanation:

given data

battery = 4 AA

battery is rated = 1.2V, 2550 mAh

power = 3 W

efficient = 95%

to find out

How long will the LED run before the batteries are depleted

solution

we consider here power delivered by battery is = x

so power = x × efficient

3 = x 95%

x = 3.15 W

and

voltage by 4 battery is = 4 × 1.2 = 4.8 V

so current will be = $\frac{x}{volatge}$

current = $\frac{3.15}{4.8}$

current = 0.6577 A

so total discharge hours = $\frac{2550}{current}$

total discharge hours = $\frac{2550}{657.7}$

time = 3.87 hours

so LED run before the batteries are depleted is 3.87 hours

5 0

3 years ago